Question

Consider
f(x) = e^(7x − 3)
Find the inverse of the function.
f^ (−1)(x) =
Would you be able to show me the steps to solve the problem? Thanks.

Answer 1

\[\Large f(x)=e^{7x-3}\]Writing f(x) as y,\[\Large y=e^{7x-3}\]Swapping x and y,\[\Large x=e^{7y-3}\]
This new \(\Large y\) in the problem represents \(\Large f^{-1}(x)\), so we want to try and solve for it.

Answer 2

The next step is a little tricky.
We want to take the natural log of both sides.\[\Large \ln(x)=\ln\left(e^{7y-3}\right)\]From there we can apply a rule of logarithms:\[\Large \color{royalblue}{\log(a^b) \quad=\quad b\cdot\log(a)}\]

Answer 3

Understand what that will do on the right side?

Answer 4

Yes, thank you.

Answer 5

lnx = 7y-(3lne)
then 1/7(lnx) = 7/7-((3lne))/7 Am I right so far?

Answer 6

Hmm your step where you pulled the exponent out of the log looks a little weird.
We want to pull it down in it's own set of brackets, like this:\[\Large \ln(e^{7x-3}) \quad=\quad (7x-3)\ln(e)\]

Answer 7

oh that's a y, not an x. my bad :)

Answer 8

(7y-3)lne = lnx
then what?
lne =lnx^(7y-3)

Answer 9

correction
lnx =lny^(7y-3)

Answer 10

From here:\[\Large \ln(e^{7x-3}) \quad=\quad (7x-3)\color{royalblue}{\ln(e)}\]
We would make note of this fact ~
If we take the log of a number which matches the base, then the result is simply 1.
Examples:
\[\Large \log_2(2)=1\]\[\Large \log_e(e)=1\]\[\Large \log_{10}(10)=1\]

Answer 11

Ahhh I meant to type,\[\Large \ln x \quad=\quad (7x-3)\color{royalblue}{\ln(e)}\]on the first line. anyway,

Answer 12

So using that information, and remembering what the natural log actually is,\[\Large \ln(e) \quad=\quad \log_e(e) \quad = \quad ?\]

Answer 13

Wow i keep screwing up the variables.. sorry about that :(\[\Large \ln x \quad=\quad (7y-3)\color{royalblue}{\ln(e)}\]